Question

In the figure angle ABC = angle ADC and AB = AD . Prove that angle BCD is an isosceles triangle?​

Answer 1

Answer:

In △ABC, we have

AB=AC ∣ given

∠ACB=∠ABC ... (1) ∣ Since angles opp. to equal sides are equal

Now, AB=AD ∣ Given

∴AD=AC ∣ Since AB=AC

Thus , in △ADC, we have

AD=AC

⇒∠ACD=∠ADC ... (2) ∣ Since angles opp. to equal sides are equal

Adding (1) and (2) , we get

∠ACB+∠ACD=∠ABC+∠ADC

⇒∠BCD=∠ABC+∠BDC ∣ Since∠ADC=∠BDC

⇒∠BCD+∠BCD=∠ABC+∠BDC+∠BCD ∣ Adding ∠BCD on both sides

⇒2∠BCD=180

∘

∣ Angle sum property

⇒∠BCD=90

∘

Hence, ∠BCD is a right angle.

Answer 2

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